Question

Let $f:R\to R$ and $g:R\to R$ be functions satisfying $f(x+y)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)$ and $f\left(x\right)=xg\left(x\right)$ for all $x,y\in R.$ If $\underset{x\to 0}{lim}g\left(x\right)=1,$ then which of the following statements is/are TRUE?

• A. $f$ is differentiable at every $x\in R$
• B. If $g\left(0\right)=1,$ then $g$ is differentiable at every $x\in R$
• C. The derivative $f^{\prime }\left(1\right)$ is equal to $1$
• D. The derivative $f^{\prime }\left(0\right)$ is equal to $1$

Answer 1

Answer: (A), (B), (D)

The derivative $f^{\prime }\left(0\right)$ is equal to $1$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+f\left(x\right)f\left(h\right)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(h\right)\left(f\right(x)+1)}{h}$
$=\underset{h\to 0}{lim}\frac{hg\left(h\right)\left(f\right(x)+1)}{h}$
$=f\left(x\right)+1(\because \underset{x\to 0}{lim}g\left(x\right)=1)$
$\therefore f^{\prime }\left(x\right)=f\left(x\right)+1$

$\frac{f^{\prime }\left(x\right)}{1+f\left(x\right)}=1$
On integrating both sides, we get $\int \frac{f^{\prime }\left(x\right)}{1+f\left(x\right)}dx=\int dx$
$\Rightarrow \ln (f\left(x\right)+1)=x+C\cdots \left(1\right)$

$f(x+h)=f\left(x\right)+f\left(h\right)+f\left(x\right)f\left(h\right)$ for all $x,h\in R$
$\Rightarrow \underset{h\to 0}{lim}f(x+h)=\underset{h\to 0}{lim}[f\left(x\right)+f\left(h\right)+f\left(x\right)f\left(h\right)]$
$\Rightarrow \underset{h\to 0}{lim}f(x+h)=f\left(x\right)+\underset{h\to 0}{lim}[hg\left(h\right)+f\left(x\right)hg\left(h\right)]$
$\Rightarrow \underset{h\to 0}{lim}f(x+h)=f\left(x\right)+0=f\left(x\right)$
Similarly, $\underset{h\to 0}{lim}f(x-h)=f\left(x\right)$
$\therefore f$ is continuous for all $x\in R$
$\Rightarrow f\left(0\right)=\underset{h\to 0}{lim}f\left(h\right)=0$

Now, putting $x=0$ in equation $\left(1\right)$
$\ln (1+f\left(0\right))=C$
$\Rightarrow C=0$
$\therefore 1+f\left(x\right)=e^x$
$f\left(x\right)=e^x-1$
$f^{\prime }\left(x\right)=e^x$
$f^{\prime }\left(1\right)=e$ and $f^{\prime }\left(0\right)=1$

$g\left(x\right)=\frac{f\left(x\right)}{x}=\{\begin{array}{cc}\frac{e^x-1}{x},& x\ne 0\\ 1,& x=0\end{array}$

We have to check differentiability at $x=0$
$g^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\left(\frac{\frac{e^h-1}{h}-1}{h}\right)$
$=\underset{h\to 0}{lim}\left(\frac{e^h-1-h}{h^2}\right)=\frac{1}{2}$
$g^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\left(\frac{\frac{e^{-h}-1}{-h}-1}{-h}\right)$
$=\underset{h\to 0}{lim}\left(\frac{e^{-h}-1+h}{h^2}\right)=\frac{1}{2}$
If $g\left(0\right)=1,$ then $g$ is differentiable at $x=0$ and hence differentiable at every $x\in R$